Let a,b be elements of a group G. show that if ab has finite order, then ba has finite order.
The Attempt at a Solution
Let n be the order of ab so that (ab)n = e. Multiplying this equation on the left by b and on the right by a, we find that (ba)n+1 = bea = (ba)e. Cancellation of the first factor ba from both sides shows that (ba)n = e, so the order of ba is < n. If the order of ba were less than n, a symmetric argument would show that the order of ab is less than n, contrary to our choice of n. Thus ba has order n also.
okay. so, this is from a chapter on cyclic groups, so i'm assuming that it has to do with, duh, cyclic groups. i know cyclic groups are abelian, and if that were to be assumed from the beginning, i believe this problem would be relatively easy. although, actually, it probably wouldn't even be a problem, since it would be true from the beginning (since we could just say ab=ba, and abn=ban
Let n be the order of ab so that (ab)n = e.
got that, that makes sense, i believe that's the definition of order of an element.
WAIT. here's the crux, something i (maybe) just realized as i typed this sentence. since we define the order of ab such that abn=e, does that imply that G is cyclic, and thus abelian? that would make things easier, if i'm right. so far, a quick internet search has failed me on an answer.